If `f(x)=e^xsinx` in `[0,\ pi]` , then `c` in Rolles theorem is `pi//6` (b) `pi//4` (c) `pi//2` (d) `3pi//4`

Explanation

To solve the problem using Rolle's Theorem for the function \( f(x) = e^x \sin x \) on the interval \([0, \pi]\), we will follow these steps:

### Step 1: Check the conditions of Rolle's Theorem
Rolle's theorem states that if a function is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\).

**Hint:** Verify the continuity and differentiability of the function on the given interval.

### Step 2: Evaluate \(f(0)\) and \(f(\pi)\)
Calculate the values of the function at the endpoints of the interval:
- \( f(0) = e^0 \sin(0) = 1 \cdot 0 = 0 \)
- \( f(\pi) = e^{\pi} \sin(\pi) = e^{\pi} \cdot 0 = 0 \)

Since \(f(0) = f(\pi)\), the condition \(f(a) = f(b)\) is satisfied.

**Hint:** Confirm that the function values at the endpoints are equal.

### Step 3: Differentiate the function
Now we need to find the derivative of \(f(x)\):
\[ f'(x) = \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x) \]

**Hint:** Use the product rule for differentiation.

### Step 4: Set the derivative equal to zero
To find \(c\), we set the derivative equal to zero:
\[ e^c (\sin c + \cos c) = 0 \]
Since \(e^c\) is never zero, we have:
\[ \sin c + \cos c = 0 \]

**Hint:** Recognize that \(e^c\) does not affect the equality since it is always positive.

### Step 5: Solve for \(c\)
Rearranging gives:
\[ \sin c = -\cos c \]
This can be rewritten as:
\[ \tan c = -1 \]
The solutions to \(\tan c = -1\) in the interval \([0, \pi]\) occur at:
\[ c = \frac{3\pi}{4} \]

**Hint:** Recall the angles where the tangent function is negative.

### Conclusion
Thus, the value of \(c\) that satisfies Rolle's Theorem for the function \(f(x) = e^x \sin x\) on the interval \([0, \pi]\) is:
\[ \boxed{\frac{3\pi}{4}} \]